The number of blocks of size at least $k$ corresponding to $\lambda$ in the Jordan canonical form is equal to nullity of $(A-\lambda I)^k$ minus the nullity of $(A-\lambda I)^{k-1}$.

The number of companion blocks in the rational canonical form corresponding to powers $p(t)^m$ with $m\geq k$ (where $p(t)$ is an irreducible factor of the minimal polynomial) is likewise equal to the nullity of $p(A)^m$ minus the nullity of $p(A)^{m-1}$.

If the minimal polynomial of $A$ splits, then the irreducible factors of the minimal polynomial are all of the form $(x-\lambda)$, so the number of companion blocks correspond to powers $(x-\lambda)^m$ with $m\geq k$ is exactly equal to the number of Jordan blocks in the Jordan canonical form of size at least $k$ and corresponding to $\lambda$.

In particular, the companion matrix of $(x-1)^n$ is already in Rational Canonical form, and has a single block corresponding to the power $(x-1)^n$; its Jordan canonical form will therefore have exactly one block, corresponding to $\lambda=1$, and of size $n\times n$. Since any matrix is similar to its Jordan canonical form, it follows that the $n\times n$ Jordan block matrix with $1$s in the diagonal is similar to the companion matrix of $(x-1)^n$.

**Added.**

To see why the Pascal matrix realizes the similarity, let $A$ be the $n\times n$ matrix with $1$s in the diagonal and $1$s directly above (i.e., a single $n\times n$ Jordan block associated to $1$). Then the standard ordered basis $[\mathbf{e}_1,\ldots,\mathbf{e}_n]$ is a Jordan canonical basis for $A$, and we have $A\mathbf{e}_1=\mathbf{e}_1$, $A\mathbf{e}_k = \mathbf{e}_{k-1}+\mathbf{e}_k$, $k=2,\ldots, n$.

To get a rational canonical basis, we need a vector $\mathbf{v}$ that lies in the kernel of $(A-I)^n$, but not in the kernel of $(A-I)^{n-1}$; then the rational canonical basis will be $[\mathbf{v},A\mathbf{v},A^2\mathbf{v},\ldots,A^{n-1}\mathbf{v}]$. It is not hard to verify that $(A-I)^k$ is a matrix that has $1$s in the $k$th diagonal above the main diagonal and $0$s elsewhere. So $(A-I)^n$ is the zero matrix, and $(A-I)^{n-1}$ is a matrix whose only nonzero entry is a $1$ in the $(1,n)$ entry. Thus, we can take $\mathbf{v}=\mathbf{e}_n$. So the rational canonical ordered basis is $[\mathbf{e}_n,A\mathbf{e}_n,\ldots,A^{n-1}\mathbf{e}_{n-1}]$, or explicitly:
$$\begin{align*}
&\phantom{=} \mathbf{e}_n\\
A\mathbf{e}_n &= \mathbf{e}_{n-1}+\mathbf{e}_n\\
A^2\mathbf{e}_n &= \mathbf{e}_{n-2} + 2\mathbf{e}_{n-1}+\mathbf{e}_n\\
A^3\mathbf{e}_n &= \mathbf{e}_{n-3}+3\mathbf{e}_{n-2}+3\mathbf{e}_{n-1}+\mathbf{e}_n\\
&\vdots\\
A^{n-1}\mathbf{e}_n &= \binom{n-1}{0}\mathbf{e}_1 + \binom{n-1}{1}\mathbf{e}_{2} + \binom{n-1}{2}\mathbf{e}_3 + \cdots + \binom{n-1}{n-2}\mathbf{e}_{n-1} + \binom{n-1}{n-1}\mathbf{e}_n
\end{align*}$$
The change-of-basis matrix from the rational canonical basis to the standard basis is therefore:
$$Q=\left(\begin{array}{ccccc}
0 & 0 &\cdots & 0 &\binom{n-1}{n-1}\\
0 & 0 & \cdots &\binom{n-2}{n-2} & \binom{n-1}{n-2}\\
0 & 0& \cdots & \binom{n-2}{n-3} & \binom{n-1}{n-3}\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
0 & \binom{1}{1} & \cdots & \binom{n-2}{1} & \binom{n-1}{1}\\
1 & \binom{1}{0} & \cdots & \binom{n-2}{0} & \binom{n-1}{0}
\end{array}\right)$$
so $Q^{-1}AQ$ will be the companion matrix of $(x-1)^n$.

**Added${}^{\mathbf{2}}$.** Note that the argument in the first paragraph shows that an $n\times n$ Jordan block associated to $\lambda$ is similar to the companion matrix of $(x-\lambda)^n$, for any $\lambda$, not just for $n=1$; and more generally, if we let $J(\lambda,k)$ denote a $k\times k$ Jordan block associated to $\lambda$; $C(p(x))$ denote the companion matrix of $p(x)$; and $B_1\oplus B_2\oplus \cdots \oplus B_n$, with $B_i$ square matrices, denote the block-diagonal matrix with diagonal blocks equal to $B_1$, $B_2,\ldots,B_n$, then the matrix
$$J(\lambda_1,k_1)\oplus J(\lambda_2,k_2)\oplus\cdots\oplus J(\lambda_m,k_m)$$
is similar to the matrix
$$C((x-\lambda_1)^{k_1}) \oplus C((x-\lambda_2)^{k_2})\oplus\cdots\oplus C((x-\lambda_m)^{k_m}),$$
with your question being the case $m=1$, $\lambda_1=1$, $k_1=n$.